Locus of middle points of the chords of conicoid which are parallel to $xy$ plane and touch the given sphere

Question

I have the following conicoid before me:

$ax^2+by^2+cz^2=1$

I have to find the locus of the middle points of the chords which are parallel to the plane $z=0$ and touch the sphere $x^2+y^2+z^2=a^2$. This is what I have tried:

Let $(A,B,C)$ denote the mid point of a chord. Let $<l,m,n>$ be the direction cosines of the chord.

Then equation of chord is given by:

$\dfrac{x-A}{l}=\dfrac{y-B}{m}=\dfrac{z-C}{n}= r$ (say)

Then $x=lr+A$;

$y=mr+B$;

$z=nr+C$

As the chords are parallel to $z=0$ plane hence $n=0$. This means that $z=C$.

Putting values of $x$,$y$ and $z$ in the equation of conicoid, we get:

$a(lr+A)^2+b(mr+B)^2+cC^2=1$ which gives:

$r^2(al^2+bm^2)+2r(alA+bmB)+aA^2+bB^2+cC^2-1=0$

For getting equal and opposite values of $r$, sum of roots should be $0$ i.e.

$alA+bmB=0$

Lastly, I tried to use the sphere condition. I again put the values of $x$,$y$, $z$ in the equation of the sphere:

$(lr+A)^2+(mr+B)^2+C^2=a^2$ which means

$r^2(l^2+m^2)+2r(lA+mB)+A^2+B^2+C^2-a^2=0$

As the chord only touches the sphere, roots must be equal which happens when:

$4(lA+mB)^2-4(l^2+m^2)(A^2+B^2+C^2-a^2)=0$ which gives

$2lmAB-l^2(B^2+C^2-a^2)-m^2(A^2+C^2-a^2)=0$

After this I do not know how to proceed. Please guide me as to how to proceed further. Any suggestions will be highly appreciated.

This question is not getting any answers. At the same time, I do not know what should I do to draw the interest of the community towards this question. Please help me out with this question. I really need to know if I am doing this problem correctly or not. And if yes, what is the way forward from the point I am stuck at.

Answer 1

I have figured out the answer on my own. Here it goes:

Dividing final equation by $lm$,we get:

$\dfrac{l}{m}.2AB -(B^2+C^2-a^2)(\dfrac{l}{m})^2-(A^2+C^2-a^2)=0$

From equation $alA+bmB=0$, we get:

$\dfrac{l}{m}=-\dfrac{bB}{aA}$

Putting this value of $\dfrac{l}{m}$ in first equation, we get:

$\dfrac{2B^2b}{a}+(B^2+C^2-a^2)\dfrac{b^2B^2}{a^2A^2}+(A^2+C^2-a^2)=0$

Finally replacing $A$,$B$ and $C$ with $x$, $y$ and $z$ respectively, we get the required locus.